1
GATE EE 2014 Set 2
+1
-0.3
The switch SW shown in the circuit is kept at position ‘1’ for a long duration. At t = 0+, the switch is moved to position ‘2’ Assuming $$\left|V_{02}\right|\;>\;\left|V_{01}\right|$$, the voltage $$V_C\left(t\right)$$ across capacitor is
A
$$v_c\left(t\right)=-V_{02}\left(1-e^{-t/2RC}\right)-V_{01}$$
B
$$v_c\left(t\right)=V_{02}\left(1-e^{-t/2RC}\right)+V_{01}$$
C
$$v_c\left(t\right)=-\left(V_{02}+V_{01}\right)\left(1-e^{-t/2RC}\right)-V_{01}$$
D
$$v_c\left(t\right)=\left(V_{02}-V_{01}\right)\left(1-e^{-t/2RC}\right)+V_{01}$$
2
GATE EE 2012
+1
-0.3
In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i(t) for all t is
A
zero
B
a step function
C
an exponentially decaying function
D
an impulse function
3
GATE EE 2010
+1
-0.3
The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = 0+, the current through the 1μF capacitor is
A
0 A
B
1 A
C
1.25 A
D
5 A
4
GATE EE 2005
+1
-0.3
In the Fig. given below, the initial capacitor voltage is zero. The switch is closed at $$t = 0.$$ the final steady-state voltage across the capacitor is
A
$$20$$ $$V$$
B
$$10$$ $$V$$
C
$$5V$$
D
$$0$$ $$V$$
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